Question

The heat liberated on complete combustion of $7.8g$ benzene is $327kJ$. This heat was measured at constant volume and at ${27}^{o}C$. Calculate the heat of combustion of benzene at constant pressure.

• A. $-3274kJmo{l}^{-1}$
• B. $-1637kJmo{l}^{-1}$
• C. $-3270kJmo{l}^{-1}$
• D. None of the above

Answer 1

Answer: (A)

$-3274kJmo{l}^{-1}$

Solution:- (A) $-3274kJ/mol$

Molecular weight of $C_6{H}_6=78g$

Heat liberated on combustion of $7.8g$ benzene $=327kJ$

Heat liberated on combustion of $78g$ benzene $=\frac{327}{7.8}\times 78=3270kJ$

$\therefore \Delta E=-3270kJ$

Here negative sign indicates that the energy is liberated.

${C_6{H}_6}_{\left(l\right)}+7\frac{1}{2}{O_2}_{\left(g\right)}⟶6{C{O}_2}_{\left(g\right)}+3{H_{2}O}_{\left(l\right)}$

From the above reaction,

$\Delta n_g=n_P-n_R=6-\frac{15}{2}=-\frac{3}{2}$

Temperature $\left(T\right)=25℃=(25+273)K=298K\left(\text{Given}\right)$

Now from first law of thermodynamics,

$\Delta H=\Delta E+\Delta n_{g}RT$

$\Delta H=\Delta E+(-\frac{3}{2})\times 8.314\times {10}^{-3}\times 300$

$\Rightarrow \Delta H=-3270-3.74=-3273.74kJ/mol\approx -3274kJ/mol$

Hence the heat of reaction at constant pressure will be $-3274kJ/mol$.